Answer: (2*p + 3)/q
Step-by-step explanation:
First, let's remember the relationships:
Logₙ(a) = Ln(a)/Ln(n)
Ln(A*B) = Ln(A) + Ln(B)
Ln(a^n) = n*Ln(a)
Now, we know that:
Logₓ(2) = p
Logₓ(7) = q
We want to express:
Log₇(4*x^3) in terms of p and q.
First, we can rewrite the first two relations as:
Ln(2)/Ln(x) = p
Ln(7)/ln(x) = q
then we have:
Ln(2) = p*Ln(x)
Ln(7) = q*Ln(x)
Ok:
Now let's play with our equation:
Log₇(4*x^3)
First, this is equal to:
Ln(4*x^3)/Ln(7)
We now can rewrite this as:
(Ln(4) + Ln(x^3))/Ln(7)
= (Ln(2^2) + Ln(x^3))/Ln(7)
= (2*Ln(2) + 3*Ln(x))/Ln(7)
Now we can replace Ln(2) by p*Ln(x) and Ln(7) by q*Ln(x)
(2*p*Ln(x) + 3*Ln(x))/(q*Ln(x)) = (2*p + 3)/q
This is the expression we wanted.
Answer:
y = -x + 5
y = x/2 + 3/2
Step-by-step explanation:
First we calculate the slope as difference between coordinates
a) P₁ ( -1 , 6 ) P₂ ( 3 , 2 )
m = Δy/Δx ⇒ m = ( y₂ - y₁ ) / ( x₂ - x₁ ) ⇒ m = (2 -6 ) / (3 -(-1))
m = -4 /4
m = -1
Now we have:
y - y₁ = m ( x - x₁ ) ⇒ y - 6 = (-1) ( x - (-1))
y - 6 = -1 * ( x + 1 ) ⇒ y - 6 = - x - 1
y = -x + 5
In the second case
Q₁ ( -3 , 0 ) Q₂ ( 5 , 4 )
Again we calculate the slope
m = Δy / Δx ⇒ m = ( 4 - 0 ) / ( 5 - (-3)) ⇒ m = 4 / 8 ⇒ m = 1/2
m = 1/2 or x = 0,5
And
y - y₁ = 1/2 ( x - x₁ ) ⇒ y - y₁ = 1/2 ( x - (-3)) ⇒ y - 0 = 1/2 * x + 3/2
y = x/2 + 3/2
Answer:
4.75% probability that the line pressure will exceed 1000 kPa during any measurement
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that the line pressure will exceed 1000 kPa during any measurement
This is 1 subtracted by the pvalue of Z when X = 1000. So
has a pvalue of 0.9525
1 - 0.9525 = 0.0475
4.75% probability that the line pressure will exceed 1000 kPa during any measurement
Answer:
3x6+3x1+4x3
Step-by-step explanation:
